Focus: Density, Buoyancy, and Archimedes' Principle
A block of aluminum has a mass of 540 g and a volume of 200 cm³. What is its density in g/cm³?
Mercury has a density of 13.6 g/cm³. What is the volume of 272 g of mercury?
The density of gold is 19.3 g/cm³. Convert this density to the SI unit of kg/m³.
An object has a density of 1200 kg/m³. Will it float or sink in fresh water (density $\approx$ 1000 kg/m³)? Explain why.
A solid iron rod is shaped like a cylinder with a radius of 2 cm and a length of 50 cm. If the density of iron is 7.87 g/cm³, what is the mass of the rod in kilograms?
A rock with a volume of 500 cm³ is dropped into a cylindrical container of water with a base radius of 10 cm. By how much does the water level rise?
A piece of granite has a mass of 2.7 kg and a density of 2700 kg/m³. It is fully submerged in water (density 1000 kg/m³). What is its apparent mass when submerged?
An alloy is made by mixing 100 cm³ of copper ($\rho = 8.96$ g/cm³) with 50 cm³ of zinc ($\rho = 7.14$ g/cm³). Assuming the final volume is the sum of the initial volumes, what is the density of the alloy?
An object in the shape of a perfect cube with side length $s = 10$ cm has a uniform density. It is placed in a container holding two immiscible liquids: oil and water. The oil layer is on top of the water. The densities are $\rho_{\text{oil}} = 800$ kg/m³ and $\rho_{\text{water}} = 1000$ kg/m³. The cube comes to rest in equilibrium, with 40% of its volume submerged in the water and the rest in the oil. Calculate the density of the cube.
A hollow sphere made of a metal with density $\rho_{\text{metal}} = 7800$ kg/m³ has an outer radius $R = 12$ cm and an inner radius $r = 10$ cm. It is fully submerged in a large tank of fresh water ($\rho_{\text{water}} = 1000$ kg/m³). Calculate the apparent weight of the sphere when it is submerged.
A simple hydrometer consists of a cylindrical glass tube of radius $r = 1$ cm, weighted at the bottom. The total mass of the hydrometer is 50 g. It is floated in a liquid of unknown density, and it settles with a length $L_1 = 20$ cm of the tube submerged. When placed in water ($\rho_{\text{water}} = 1.00$ g/cm³), it settles with $L_2 = 25$ cm submerged. What is the density of the unknown liquid?
A block of copper has a density of $\rho_0 = 8960$ kg/m³ at a temperature $T_0 = 20^\circ$C. The coefficient of linear expansion for copper is $\alpha = 17 \times 10^{-6}$ (C$^\circ$)$^{-1}$. What is the density of the copper block when it is heated to a temperature of $T_f = 520^\circ$C?
The formula for density is $\rho = \frac{m}{V}$.
$\rho = \frac{540 \text{ g}}{200 \text{ cm³}} = 2.7$ g/cm³.
Rearranging the density formula, $V = \frac{m}{\rho}$.
$V = \frac{272 \text{ g}}{13.6 \text{ g/cm³}} = 20$ cm³.
To convert from g/cm³ to kg/m³, we multiply by 1000.
Explanation: $1 \frac{\text{g}}{\text{cm³}} = 1 \frac{\text{g}}{\text{cm³}} \times \frac{1 \text{ kg}}{1000 \text{ g}} \times (\frac{100 \text{ cm}}{1 \text{ m}})^3 = \frac{1}{1000} \times 1,000,000 \frac{\text{kg}}{\text{m³}} = 1000 \frac{\text{kg}}{\text{m³}}$.
$19.3 \text{ g/cm³} \times 1000 = 19300$ kg/m³.
The object will sink. An object sinks if its density is greater than the density of the fluid. Since $1200 \text{ kg/m³} > 1000 \text{ kg/m³}$, the object is denser than water and will sink.
1. Find the volume of the cylinder: $V = \pi r^2 h$.
$V = \pi (2 \text{ cm})^2 (50 \text{ cm}) = 200\pi \approx 628.32$ cm³.
2. Find the mass: $m = \rho \times V$.
$m = 7.87 \text{ g/cm³} \times 628.32 \text{ cm³} \approx 4944.9$ g.
3. Convert to kilograms: $4944.9 \text{ g} \div 1000 = 4.94$ kg.
The volume of water displaced is equal to the volume of the rock, $V_{\text{displaced}} = 500$ cm³.
This displaced volume takes the shape of a cylinder with the same base area as the container. The volume of this shape is $V = A_{\text{base}} \times \Delta h$.
$A_{\text{base}} = \pi r^2 = \pi (10 \text{ cm})^2 = 100\pi$ cm².
$\Delta h = \frac{V_{\text{displaced}}}{A_{\text{base}}} = \frac{500 \text{ cm³}}{100\pi \text{ cm²}} \approx 1.59$ cm. The water level rises by about 1.59 cm.
1. Find the volume of the granite: $V = \frac{m}{\rho} = \frac{2.7 \text{ kg}}{2700 \text{ kg/m³}} = 0.001$ m³.
2. Calculate the buoyant force: $F_B = \rho_{\text{water}} \cdot V \cdot g = 1000 \text{ kg/m³} \cdot 0.001 \text{ m³} \cdot 9.8 \text{ m/s²} = 9.8$ N.
3. Calculate the actual weight: $W = m \cdot g = 2.7 \text{ kg} \cdot 9.8 \text{ m/s²} = 26.46$ N.
4. Find the apparent weight: $W_{\text{app}} = W - F_B = 26.46 \text{ N} - 9.8 \text{ N} = 16.66$ N.
5. Find the apparent mass: $m_{\text{app}} = \frac{W_{\text{app}}}{g} = \frac{16.66 \text{ N}}{9.8 \text{ m/s²}} = 1.7$ kg.
1. Find the total mass: $m_{\text{total}} = m_{\text{copper}} + m_{\text{zinc}}$.
$m_{\text{copper}} = \rho_{Cu} V_{Cu} = 8.96 \text{ g/cm³} \times 100 \text{ cm³} = 896$ g.
$m_{\text{zinc}} = \rho_{Zn} V_{Zn} = 7.14 \text{ g/cm³} \times 50 \text{ cm³} = 357$ g.
$m_{\text{total}} = 896 \text{ g} + 357 \text{ g} = 1253$ g.
2. Find the total volume: $V_{\text{total}} = 100 \text{ cm³} + 50 \text{ cm³} = 150$ cm³.
3. Calculate alloy density: $\rho_{\text{alloy}} = \frac{m_{\text{total}}}{V_{\text{total}}} = \frac{1253 \text{ g}}{150 \text{ cm³}} \approx 8.35$ g/cm³.
Principle: For an object in equilibrium (floating), its total weight must be balanced by the total upward buoyant force from the fluids it displaces.
Weight of cube: $W = m_{\text{cube}} \cdot g = \rho_{\text{cube}} \cdot V_{\text{total}} \cdot g$
Buoyant force from oil: $F_{B,\text{oil}} = \rho_{\text{oil}} \cdot V_{\text{oil}} \cdot g$
Buoyant force from water: $F_{B,\text{water}} = \rho_{\text{water}} \cdot V_{\text{water}} \cdot g$
Setup: According to Archimedes' principle: $W = F_{B,\text{oil}} + F_{B,\text{water}}$.
$\rho_{\text{cube}} V_{\text{total}} g = \rho_{\text{oil}} V_{\text{oil}} g + \rho_{\text{water}} V_{\text{water}} g$
We can cancel $g$ from all terms: $\rho_{\text{cube}} V_{\text{total}} = \rho_{\text{oil}} V_{\text{oil}} + \rho_{\text{water}} V_{\text{water}}$
Calculation: Given that 40% of the volume is in water, $V_{\text{water}} = 0.40 V_{\text{total}}$. This means the remaining 60% is in oil, so $V_{\text{oil}} = 0.60 V_{\text{total}}$.
$\rho_{\text{cube}} V_{\text{total}} = \rho_{\text{oil}} (0.60 V_{\text{total}}) + \rho_{\text{water}} (0.40 V_{\text{total}})$
Divide the entire equation by $V_{\text{total}}$: $\rho_{\text{cube}} = 0.60 \cdot \rho_{\text{oil}} + 0.40 \cdot \rho_{\text{water}}$
$\rho_{\text{cube}} = 0.60 \cdot (800 \text{ kg/m³}) + 0.40 \cdot (1000 \text{ kg/m³})$
$\rho_{\text{cube}} = 480 + 400 = 880$ kg/m³
Final Answer: The density of the cube is 880 kg/m³.
Principle: The apparent weight of a submerged object is its actual weight in air minus the upward buoyant force.
Apparent Weight: $W_{\text{app}} = W_{\text{actual}} - F_{B}$
1. Calculate the volume of the metal: This is the volume of the outer sphere minus the volume of the inner hollow part.
$V_{\text{metal}} = V_{\text{outer}} - V_{\text{inner}} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (R^3 - r^3)$
Convert radii to meters: $R = 0.12$ m, $r = 0.10$ m.
$V_{\text{metal}} = \frac{4}{3}\pi ((0.12)^3 - (0.10)^3) = \frac{4}{3}\pi (0.001728 - 0.001) \approx 0.00305$ m³
2. Calculate the actual weight: $W_{\text{actual}} = m_{\text{sphere}} \cdot g = (\rho_{\text{metal}} \cdot V_{\text{metal}}) \cdot g$
$W_{\text{actual}} = (7800 \text{ kg/m³} \cdot 0.00305 \text{ m³}) \cdot 9.8 \text{ m/s²} \approx 233.1$ N
3. Calculate the buoyant force: The buoyant force depends on the total volume of water displaced, which corresponds to the sphere's *outer* volume.
$F_B = \rho_{\text{water}} \cdot V_{\text{displaced}} \cdot g = \rho_{\text{water}} \cdot (\frac{4}{3}\pi R^3) \cdot g$
$F_B = 1000 \text{ kg/m³} \cdot (\frac{4}{3}\pi (0.12)^3) \cdot 9.8 \text{ m/s²} \approx 70.9$ N
4. Calculate the apparent weight: $W_{\text{app}} = 233.1 \text{ N} - 70.9 \text{ N} = 162.2$ N
Final Answer: The apparent weight of the submerged sphere is 162.2 N.
Principle: A floating object displaces a weight of fluid equal to its own weight. Since the hydrometer's weight is constant, the weight of the displaced fluid is the same in both the unknown liquid and in water.
$W_{\text{hydrometer}} = W_{\text{displaced, unknown}} = W_{\text{displaced, water}}$
Setup: Let $\rho_{\text{unk}}$ be the density of the unknown liquid. The displaced volume is the cylinder's cross-sectional area ($A$) times the submerged length ($L$).
$(\rho_{\text{unk}} \cdot A \cdot L_1) \cdot g = (\rho_{\text{water}} \cdot A \cdot L_2) \cdot g$
Calculation: We can cancel the area $A$ and gravity $g$ from both sides, as they are constant.
$\rho_{\text{unk}} \cdot L_1 = \rho_{\text{water}} \cdot L_2$
$\rho_{\text{unk}} = \rho_{\text{water}} \cdot \frac{L_2}{L_1}$
$\rho_{\text{unk}} = (1.00 \text{ g/cm³}) \cdot \frac{25 \text{ cm}}{20 \text{ cm}} = 1.25$ g/cm³
Final Answer: The density of the unknown liquid is 1.25 g/cm³.
Principle: When an object is heated, its volume increases while its mass remains constant. Since density $\rho = m/V$, an increase in volume causes a decrease in density.
1. Relate volume expansion to linear expansion: For isotropic materials, the coefficient of volume expansion, $\beta$, is approximately three times the coefficient of linear expansion, $\alpha$.
$\beta \approx 3\alpha = 3 \cdot (17 \times 10^{-6}) = 51 \times 10^{-6}$ (C$^\circ$)$^{-1}$
2. Find the new volume: The final volume $V_f$ is related to the initial volume $V_0$ by $V_f = V_0(1 + \beta \Delta T)$.
$\Delta T = T_f - T_0 = 520^\circ\text{C} - 20^\circ\text{C} = 500^\circ$C
$V_f = V_0(1 + (51 \times 10^{-6}) \cdot 500) = V_0(1 + 0.0255) = 1.0255 V_0$
3. Find the new density: The mass $m$ is constant. $\rho_f = m/V_f$ and $\rho_0 = m/V_0$.
$\rho_f = \frac{m}{1.0255 V_0} = \frac{1}{1.0255} \cdot \frac{m}{V_0} = \frac{\rho_0}{1.0255}$
$\rho_f = \frac{8960 \text{ kg/m³}}{1.0255} \approx 8737.2$ kg/m³
Final Answer: The new density of the copper is approximately 8737.2 kg/m³.